LeetCode in Kotlin
1855. Maximum Distance Between a Pair of Values
Medium
You are given two non-increasing 0-indexed integer arrays nums1 and nums2.
A pair of indices (i, j), where 0 <= i < nums1.length and 0 <= j < nums2.length, is valid if both i <= j and nums1[i] <= nums2[j]. The distance of the pair is j - i.
Return the maximum distance of any valid pair (i, j). If there are no valid pairs, return 0.
An array arr is non-increasing if arr[i-1] >= arr[i] for every 1 <= i < arr.length.
Example 1:
Input: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
Output: 2
Explanation: The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4). The maximum distance is 2 with pair (2,4).
Example 2:
Input: nums1 = [2,2,2], nums2 = [10,10,1]
Output: 1
Explanation: The valid pairs are (0,0), (0,1), and (1,1). The maximum distance is 1 with pair (0,1).
Example 3:
Input: nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]
Output: 2
Explanation: The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4). The maximum distance is 2 with pair (2,4).
Constraints:
1 <= nums1.length, nums2.length <= 1051 <= nums1[i], nums2[j] <= 105- Both
nums1andnums2are non-increasing.
Solution
class Solution {
fun maxDistance(nums1: IntArray, nums2: IntArray): Int {
val n = nums1.size
val m = nums2.size
var po1 = 0
var po2 = 0
var res = 0
while (po1 < n && po2 < m) {
if (nums1[po1] > nums2[po2]) {
po1++
} else {
if (po2 != po1) {
res = Math.max(res, po2 - po1)
}
po2++
}
}
return res
}
}